Question

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_4\left(x\right)$

$\sin x$

Short answer

The required answer is$R_4\left(x\right)=\frac{\cos c}{120}{x}^5$

Step-by-step solution

Step 1. Given Information  

The given function is$f\left(x\right)=\sin x$

Step 2. Explanation 

Using the Lagrange form for the remainder, we have,

$$\begin{gathered} R_n\left(x\right)=\frac{f^{n+1}\left(c\right)}{(n+1)!}{x}^{n+1} \\ {R}_4\left(x\right)=\frac{f^5\left(c\right)}{5!}{x}^5 \end{gathered}$$

Now, we will find the fifth derivative of the function as follows,

$$\begin{gathered} f^1\left(x\right)=\cos x \\ {f}^2\left(x\right)=-\sin x \\ {f}^3\left(x\right)=-\cos x \\ {f}^4\left(x\right)=\sin x \\ {f}^5\left(x\right)=\cos x \end{gathered}$$

Thus, we get,

$$\begin{gathered} R_4\left(x\right)=\frac{\cos c}{5!}{x}^5 \\ {R}_4\left(x\right)=\frac{\cos c}{120}{x}^5 \end{gathered}$$